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$f(x)= \dfrac{1}{(1-x)^2}$ on $]-\infty, 1[$.
$f$ is convex because
\[ f''(x) = \frac{6}{(1-x)^4} \geq 0. \]
And the tangent in 0 is such that
\[ y=f'(0)(x-0) + f(0) \text{ with } f'(0)=2 \text{ and } f(0)=1, \]
So we can say that, for all $x \in ]-\infty, 1[$,
$f(x) \geq 2x+1$.

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